Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Work and kinetic energy?

  1. Jul 16, 2006 #1
    work and kinetic energy????

    i was wondering if anyone could help with this proplem:

    A skier pulled by a tow rope up a frictionless ski slope that makes an angle of 12° with the horizontal. The rope moves parallel to the slope with a constant speed of 1.0 m/s. The force of the rope does 930 J of work on the skier as the skier moves a distance of 7.2 m up the incline.

    If the rope moved with a constant speed of 2.0 m/s, how much work would the force of the rope do on the skier as the skier moved a distance of 7.2 m up the incline?

    At what rate is the force of the rope doing work on the skier when the rope moves with a speed of 1.0 m/s

    At what rate is the force of the rope doing work on the skier when the rope moves with a speed of 2.0 m/s?


    all that i have is w=f*d*cos(theta) please help
     
  2. jcsd
  3. Jul 16, 2006 #2
    ok i tried sumation of work=1/2mv(final)^2-1/2mv(inital)^2. no go. then i tried to used w=fdcos(theata) 930=f*7.2*cos(12). nothing. does speed matter when dealing with force and work?!?
     
  4. Jul 16, 2006 #3

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The formula w=fdcos(theta) is perfectly fine as well as sum of the work = final kinetic energy - initial kinetic energy.

    so the work done by the rope is T d cos (theta) where T is the tension in the rope and theta is the angle between the rope and the direction of the tension, right? So the question becomes this: if the skier is pulled at constant speed at 1 m/s or at 2 m/s, what does it imply for the tension in th erope? . Is it the same at 1m/s than it is at 2 m/s? That will andwer your question.

    For the power (or rate of work) of a force, you must divide the work done by the time during which the force was in action.
     
  5. Jul 17, 2006 #4
    thank you very much
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook